HW6solutions

# HW6solutions - The 30 H inductor is in parallel with a...

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ECE35 Win’09 HW 6 Solutions Chapter 7: 7.4-4, 7.4-8, 7.5-5, 7.5-12, 7.5-13, 7.7-5, 7.8-1, 7.8-7 P7.4-4 Replacing series and parallel capacitors by equivalent capacitors, the circuit can be reduced as follows: Then P7.4-8 (a) The energy stored in the 60 mF capacitor is and the energy stored in the 20 mF capacitor is . (b) One second after the switch opens, the voltage across the capacitors is . Then and Next (c) (d) P7.5-5

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P7.5-12 Use KVL to get P7.5-13 We’ll write and solve a mesh equation. Label the meshes as shown. Apply KVL to the center mesh to get Then P7.7-5 The 25 H inductor is in series with a parallel combination of 20 H and 60 H inductors. The inductance of the equivalent inductor is

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Unformatted text preview: The 30 H inductor is in parallel with a short circuit, which is equivalent to a short circuit. After making these simplifications, we have Then P7.8-1 Then Next P7.8-7 The capacitor voltage and inductor current don’t change instantaneously and so are the keys to solving this problem. Label the capacitor voltage and inductor current as shown. Before t = 0, with the switch open and the circuit at steady state, the inductor acts like a short circuit and the capacitor acts like an open circuit. The capacitor voltage and inductor current don’t change instantaneously so and After the switch closes the circuit looks like this: From KCL: From KVL: At...
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HW6solutions - The 30 H inductor is in parallel with a...

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